# Find the second order derivatives of the function.

Question:

Find the second order derivatives of the function.

$e^{6 x} \cos 3 x$

Solution:

Let $y=e^{6 x} \cos 3 x$

Then,

$\frac{d y}{d x}=\frac{d}{d x}\left(e^{6 x} \cdot \cos 3 x\right)=\cos 3 x \cdot \frac{d}{d x}\left(e^{6 x}\right)+e^{6 x} \cdot \frac{d}{d x}(\cos 3 x)$

$=\cos 3 x \cdot e^{6 x} \cdot \frac{d}{d x}(6 x)+e^{6 x} \cdot(-\sin 3 x) \cdot \frac{d}{d x}(3 x)$

$=6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x$   ..(1)

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x\right)=6 \cdot \frac{d}{d x}\left(e^{6 x} \cos 3 x\right)-3 \cdot \frac{d}{d x}\left(e^{6 x} \sin 3 x\right)$

$=6 \cdot\left[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x\right]-3 \cdot\left[\sin 3 x \cdot \frac{d}{d x}\left(e^{6 x}\right)+e^{6 x} \cdot \frac{d}{d x}(\sin 3 x)\right] \quad[$ Using $(1)]$

$=36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-3\left[\sin 3 x \cdot e^{6 x} \cdot 6+e^{6 x} \cdot \cos 3 x \cdot 3\right]$

$=36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-18 e^{6 x} \sin 3 x-9 e^{6 x} \cos 3 x$

$=27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x$

$=9 e^{6 x}(3 \cos 3 x-4 \sin 3 x)$